50 grams salt
Volume of 50 Grams of Salt
50 Grams of Salt =
2.78 Tablespoons
8.33 Teaspoons
0.17 U.S. Cups
0.14 Imperial Cups
Answer:
The fox who shed their heavy coat in the summer.
Explanation:
If a fox has the ability to shed its coat in the summer, it will have no difficulties adapting to new temperatures even during winter. However, the heavy-coated one will.
Answer:
Wouldn't the Earth's atmosphere be moving too fast that it eventually breaks out?
Explanation:
Do NOT trust me.
Explanation:
Ionization equation for
is as follows.

s s s
Now, the expression for the solubility product is as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 
= 
As the concentration of
is given as 0.4 M.
So,
= 0.4 M
Putting the given values as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.4](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.4)
= 12.325 \times 10^{-5}[/tex]
Hence, the solubility of
in
is
.
Therefore, solubility of
in g/ml as follows.

= 0.0167 g/L
Thus, we can conclude that solubility of
is 0.0167 g/L.
Answer:
34.9 mL
Explanation:
First we <u>convert 23.6 g of LiBr into moles</u>, using its <em>molar mass </em>(86.845 g/mol):
- 23.6 g ÷ 86.845 g/mol = 0.272 mol LiBr
Now we can <u>calculate the required volume</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
- liters = moles / Molarity
- 0.272 mol / 7.8 M = 0.0349 L
We can <u>convert L into mL</u>:
- 0.0348 L * 1000 = 34.9 mL